If $x^{2} - x + a - 3 < 0$ for at-least one negative value of $x$ , then the complete set of values of $a$ is

(- \infty, 4)

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(- \infty, 2)

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(- \infty, 3)

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(- \infty, 1)

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Solution

The correct option is D $(- \infty, 3)$
Least non negative value of $x$ lies between
So, $(x) (x - 1) < 0$
$x^{2} - x < 0$
$x^{2} - x + a - 3 < 0$
$a - 3 < 0$
$a < 3$
$a \in (- a, 3)$

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